Heat Exchangers - Introduction Concentric Pipe Heat Exchange T h1 T c1 T c2 T h1 Energy Balance on Cold Stream (differential) dq C = wc p C dt C = C C dt C Energy Balance on Hot Stream (differential) dq H = wc p H dt H = C H dt H Overall Energy Balance (differential) For an adiabatic heat exchanger, the energy lost to the surroundings is zero so what is lost by one stream is gathered by the other. dq C + dq H = 0 ChE 333 1
Heat Exchange Equation It follows that the heat exchange from the hot to the cold is expressed in terms of the temperature difference between the two streams. dq H = U T H T C da The proportionality constant is the Overall heat transfer coefficient ( discussion later) Solution of the Energy Balances The Energy Balance on the two streams provides a delation for the differential temperature change. dt H = dq H and dt C C = dq C H C C However, we should recall that we have an adiabatic heat exchanger so that d T = dq H C H 1 + C H C C Overall Energy balances on each stream Hot Fluid Q H = C H T H1 T H2 Cold fluid Q C = C C T C2 T C1 Overall Energy balance on the Exchanger Q C + Q H = 0 ChE 333 2
The equation for T can be modified using the overall energy balances to yield d T = dq H T 2 C H T H1 T H2 The denominator is the energy lost by the hot stream, so d T = dq H Q H T 2 Application of the relation for energy transfer between the two streams yields d T = UdA T Q H T 2 Integration of the relation is the basis of a design equation for a heat exchanger. ln T 2 = UA Q H T 2 Rearrangement of the equation leads to The Design Equation for a Heat Exchanger Q H = UA T 2 ln T 2 = UA T lm ChE 333 3
Design of a Parallel Tube Heat Exchanger The Exchanger T h1 T c1 T c2 T h1 The Design Equation for a Heat Exchanger Q H = UA T 2 ln T 2 = UA T lm Glycerin-water solution with a Pr = 50 (at 70 C) flows through a set of parallel tubes that are plumbed between common headers. We must heat this liquid from 20 C to 60 C with a uniform wall temperature of 100 C. The flow rate, F, is 0.002 m 3 /sec (31.6 gal/sec.). How many parallel tubes are required? How do we select L and D for these tubes? Data The heat capacity, C p, is 4.2 kj/kg- K The density, ρ, is 1100 kg/m 3 The liquid has a kinematic viscosity, ν = 10 3 cm 2 /sec. ChE 333 4
Step 1 Calculate the heat load Q c = ρfc p T out T out Q c = 1100 kg m 3 0.002 sec m3 4.2 kj kg K 1 K 1 C 60 20 C Q c = 369.6 kj sec = 369.6 kwatts Step 2 Calculate the heat transfer coefficient If the flow is laminar, likely since glycerin is quite viscous, and the Re < 2000 the Nusselt number relation for laminar flow can be expressed as The Graetz number is Nu = 3.66 3 + 1.61 3 Gz 1 /3 Gz = Re Pr D L If the flow is turbulent (Re > 2000), the Nusselt numberr is given by Nu = 0.023 Re 0.8 Pr 0.4 We do not know the flow per tube and therefore we do not know the Re. However we don t need to know that. In Lecture 27 we observed for Heat Transfer in a Tube that T T R = exp πdhz T 1 T R wc p = exp 4St z D ChE 333 5
The definition of the Stanton Number is : h St = ρc p U = Nu RePr = Nu Pe Given a Re and Pr, we can calculate the Nu and the Stanton Number, the latter prviding us with the temperature at length L from the previous equation. Let s examine several configurations at L/D = 50, 100, 200. The Excel table below can be used to specify a design chart. Design Chart Pr = 50 L/D = 50 Re Nu St θcm 1 3.7610 7.52E-02 0.0233 3 3.9482 2.63E-02 0.2682 6 4.1996 1.40E-02 0.4966 10 4.4940 8.99E-03 0.6380 20 5.0980 5.10E-03 0.7750 30 5.5852 3.72E-03 0.8301 100 7.7548 1.55E-03 0.9254 200 9.5962 9.60E-04 0.9532 500 12.8779 5.15E-04 0.9746 1000 16.1628 3.23E-04 0.9840 2000 20.3244 2.03E-04 0.9899 5000 100.1133 4.00E-04 0.9802 10000 174.3074 3.49E-04 0.9827 20000 303.4868 3.03E-04 0.9849 30000 419.7714 2.80E-04 0.9861 To obtain the numbers in the spreadsheet, we used the Nusselt number relation for laminar flow expressed as Nu = 3.66 3 + 1.61 3 Gz 1 /3 ChE 333 6
and for turbulent flow as Nu = 0.023 Re 0.8 Pr 0.4 Design Chart 1.0000 0.1000 Temperature 0.0100 0.0010 0.0001 0.0000 0.0000 0.0000 L/D = 50 L/D = 100 L/D = 200 1 10 100 1000 10000 Reynolds' Number ChE 333 7
Step 3 Calculate the Area required Base case D = 2 cm. and L = 100 D = 2 meters For this case we observe that from the calculations for θ cm Reduced Temperature Re L/D = 50 L/D = 100 L/D = 200 1 0.0233 0.0006 0.0000 3 0.2682 0.0789 0.0069 8.8 0.5000 0.3966 0.1718 10 0.6380 0.4387 0.2099 12 0.6800 0.4966 0.2682 12.3 0.6854 0.5042 0.2763 24.4 0.8040 0.6836 0.5017 50 0.8805 0.8073 0.6888 60 0.8945 0.8301 0.7254 70 0.9050 0.8473 0.7532 100 0.9254 0.8805 0.8073 200 0.9532 0.9254 0.8805 500 0.9746 0.9596 0.9358 1000 0.9840 0.9746 0.9596 2000 0.9899 0.9840 0.9746 5000 0.9802 0.9913 0.9862 6000 0.9809 0.9923 0.9878 8000 0.9819 0.9936 0.9899 9000 0.9824 0.9941 0.9907 We can observe that the flow rate per tube is given by F nt so that the Reynolds number is Re = = F n t 4F πdυn t ChE 333 8
As a consequence we can observe that the total length of tubing is not dependent on D alone but on othere considerations that might set a condition for Re, e.g. a pressure drop limitation. Wv find that for this base case, we find We find that Θ cm = 0.5 n t L = A = 4F πυre L D L/D Re n t L n t 50 8.8 14.47 14.46 100 12.3 20.70 10.35 200 24.4 20.87 5.21 Does it make sense? ChE 333 9
Maximum Cooling Capacity of an Exchanger of Fixed Area Water is available for use as a coolant for an oil stream in a double-pipe heat exchanger. The flow rate of the water is 500 lb m /hr. The heat exchanger has an area of 15 ft 2. The oil heat capacity, C po, is 0.5 BTU/lb- F The overall heat transfer coefficient, U, is 50 BTU/hr-ft 2 - F The initial temperature of the water, T w0, is 100 F The maximum temperature of the water is 210 F The initial temperature of the oil, T w0, is 250 F The minimum temperature of the oil, T w0, is 140 F Estimate the maximum flow rate of oil that may be cooled assuming a fixed flow rate of water at 500 lb m /hr There are two possible modes of operation Co-current flow Counter-current flow Let us look at both cases Co-current flow Constraints T w < 210 ; T w < T o ; T o 140 Energy balances Oil Q o = F o C po T o1 T o2 = F o 0.5 250 T o2 Water Q w = F w C pw T w1 T w2 F o C p0 T o1 T o2 = 500(1.0)(210 100) = 55,000 BTU / hr ChE 333 10
Recall the Design equation Q H = UA T 2 ln T 2 = UA T lm Now the T lm is given by T lm = T 2 ln T = 2 Q w UA = 55000 (50)(15) = 73.3 Using the temperatures, we obtain T 0max = 238.5 F and from the heat balance for oil, we obtain F o = C po Q o T o1 T o2 = 0.5 250 238.5 55000 = 9560 lb / h Counter-current Flow Constraints T w < 210 ; T w < T o ; T o 140 Energy balances Oil Q o = F o C po T o1 T o2 = F o 0.5 250 T o2 Water Q w = F w C pw T w1 T w2 ChE 333 11
F o C p0 T o1 T o2 = 500(1.0)(210 100) = 55,000 BTU / hr Recall the Design equation Q H = UA T 2 ln T 2 = UA T lm Now the T lm is given by T lm = T 2 ln T = 2 Q w UA = 55000 (50)(15) = 73.3 Using the temperatures, we obtain T 0max = 221 F and from the heat balance for oil, we obtain the oil flow rate as 3800 lbm/hr. I thought that countercurrent flow was supposed to be more efficient. What is the problem? ChE 333 12